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Vinil7 [7]
2 years ago
14

A wave travels through a medium at 251 m/s and has a wavelength of 5.10 cm. What is its frequency? What is its angular frequency

? What is its period of oscillation?
Physics
1 answer:
allochka39001 [22]2 years ago
5 0

Explanation:

It is given that,

Speed of a wave, v = 251 m/s

Wavelength of the wave, λ = 5.1 cm = 0.051 m

(1) The frequency of the wave is given by :

\nu=\dfrac{v}{\lambda}

\nu=\dfrac{251\ m/s}{0.051\ m}

\nu=4921.56\ Hz

(2) Angular frequency of the wave is given by :

\omega=2\pi\nu

\omega=2\pi\times 4921.56\ Hz

\omega=30923.07\ rad/s

(3) The period of oscillation is given by T as :

T=\dfrac{1}{\nu}

T=\dfrac{1}{4921.56}

T = 0.000203 seconds

or

T = 0.203 milliseconds

Hence, this is the required solution.

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Listed following are three possible models for the long-term expansion (and possible contraction) of the universe in the absence
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Answer:

1) Recollapsing universe

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1) Recollapsing universe -in this, metric expansion of space is reverse and universe recollapses.

2) critical universe - in this, expansion of universe is very low.

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6 0
2 years ago
A wave is propagating from left to right in a medium. The particles in the medium are also vibrating from left to right. What ki
Anna71 [15]
Based on the direction of propagation compared to direction of vibration, waves are classified into:
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For the question we have here, since the direction of the wave is the same as the direction of vibration of particles, therefore, this wave is a longitudinal wave
6 0
2 years ago
Materials have unique properties because each one is made up of different kinds of which particle?
inysia [295]

D. Atoms.

Explanation:

All the matter is made of elementary particles called "atoms".

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7 0
2 years ago
Read 2 more answers
A simple pendulum 0.64m long has a period of 1.2seconds. Calculate the period of a similar pendulum 0.36m long in the same locat
weqwewe [10]

The period of the second pendulum is 0.9 s

Explanation:

The period of a simple pendulum is given by the equation

T=2\pi \sqrt{\frac{L}{g}}

where

L is the length of the pendulum

g is the acceleration of gravity at the location of the pendulum

For the first pendulum, we have

L = 0.64 m

T = 1.2 s

Therefore we can find the value of g at that location:

g=(\frac{2\pi}{T})^2 L=(\frac{2\pi}{1.2})^2 (0.64)=17.5 m/s^2

Now we can find the period of the second pendulum at the same location, which is given by

T=2\pi \sqrt{\frac{L}{g}}

where we have

L = 0.36 m (length of the  second pendulum)

g=17.5 m/s^2

Substituting,

T=2\pi \sqrt{\frac{0.36}{17.5}}=0.9 s

#LearnwithBrainly

8 0
2 years ago
A particle with a charge of 3.00 elementary charges moves through a potential difference of 4.50 volts. What is the change in el
ollegr [7]
The answer:
the relationship between elementary charge, potential difference and electrical potential energy is given by 
   E= qV
E:  lectrical potential energy
q:   elementary charge
V:   potential difference

but we have  e=abs val(q)=3 
so we have E= qV=3ex4.5V=<span>13.5 eV
</span>
the answer is <span>(4)13.5 eV</span>

8 0
2 years ago
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